JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Semiconductors

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A message signal of frequency 100 MHz and peak voltage 100 V is used to execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage 400 V. The modulation index and difference between the two side band frequencies are:

(A) $4; 2 \times 1 0^{8} Hz$
(B) $4; 1 \times 1 0^{8} Hz$
(C) $0.25; 1 \times 1 0^{8} Hz$
(D) $0.25; 2 \times 1 0^{8} Hz$ ✓ Correct

Answer: (D)

Solution

The modulation index is just the ratio of the message amplitude to the carrier amplitude,

μ = \frac{( V _{m} ) _{0}}{( V _{c} ) _{0}} = \frac{100}{400} = 0.25

. In amplitude modulation the two sidebands sit at

f_{c} + f_{m}

and

f_{c} - f_{m}

, so the gap between them is

(f_{c} + f_{m}) - (f_{c} - f_{m}) = 2 f_{m} = 2 \times 1 0^{8}

Hz. The modulation index is

0.25

and the sidebands are separated by

2 \times 1 0^{8}

Hz.

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Semiconductors

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